Does $A\in B$ imply $A\subset B$? Question: Does $A\in B$ imply $A\subset B$ and does $A\in B$ and $B\in C$ imply $A\in C$?
I've been trying to find examples to get some intuition for this and I've come up with the following:
Example 1: Suppose that $A = \{1\}$, $B = \{\{1\},2\}$. I'd say that $A$ is an element of $B$ and $A$ is a subset of $B$.
Example 2: Suppose that $A = \{1\}$, $B = \{\{1\},2\}$ and $C = \{\{\{1\},2\},3\}$. Now $A\in B$, $B\in C$ but $A\not\in C$, right?
I think my confusion stems from the fact that I'm not sure how $B = \{\{1\},2\}$ vs $B = \{1,2\}$ determines whether $A$ is an element and/or a subset of $B$.
 A: No : if $A = \{1\}$, $B = \{\{1\},2\}$, then $A\in B$ but $A\not\subset B \ $ ; all the subsets of $B$ are : $\emptyset,\ \{\{1\}\},\ \{\{2\}\},\ \{\{1\},2\}$ (moreover $B$ can't have more than $4$ subsets!).
For example 2 you are right.
A: Not all sets satisfy this property; in fact, this property is the definition of a transitive set.
Neither of your examples exhibit are transitive. In Example 1, notice that $1 \in A$ but $1 \not\in B$ (even though $\{1\} \in B$), so that $A \not\subseteq B$. In Example 2, you run into more or less the same problem.
Examples of transitive sets include the von Neumann ordinals and the sets in the cumulative hierarchy $V_{\alpha}$.
A: No. Taking $A=\{\emptyset\}$ and $B=\{A\} = \{\{\emptyset\}\}$, you have $A\in B$, but not $A\subseteq B$.

Furthermore, in Example $1$, $A=\{1\}$ is not a subset of the set $B=\{\{1\},2\}$, because it is not true that every element of $A$ is also an element of $B$. Specifically, $1$ is an element of $A$, but $1$ is not an element of $B$.
$B$ is a set with two elements, one of them is equal to $2$, the other one is equal to $\{1\}$, and $1$ is not equal to any of these two elements.
A: Your question gives an occasion to emphasize the distinction between : 
(1) not implying that proposition P is true 
and 
(2) implying that proposition P is false ( not-true). 
The fact that A is a member of B certainly does not imply that A is included in B. 
Thousands of examples could be found. 
But, precisely, due to the fact that this non-implication is so often  confirmed, we tend to think that : "A belongs to B" implies that "A is not a subset of B". 
And surprisingly, when one discovers the construction of natural numbers, one has to recognize that this induction is false. 
For  example :  2 is defined as {0,1} and 3 is defined as {0,1,2}. The number 2 is a member of the number 3. But is as also a subset of 3.  
